# Class 12 RD Sharma Solutions – Chapter 23 Algebra of Vectors – Exercise 23.7

• Last Updated : 11 Feb, 2021

### Question 1. Show that the points A, B, C with position vectors  and  are collinear.

Solution:

Given that,

Position vector of

Position vector of

Position vector of

= Position vector B – Position vector of A

= Position vector of C – Position vector of B

Using  and , we get,

So,|| but  is a common vector.

Hence, proved that A, B, C are collinear.

### Question 2 (i). If  are non-coplanar vectors, prove that the points having the position vectors  are collinear.

Solution:

Let us assume three points that are A, B, C

Position vector of A =

Position vector of B =

Position vector of C =

= Position vector of B – Position vector of A

-(1)

= Position vector of C – Position vector of B

-(2)

Using equation(1) and (2), we get

= λ           -(where λ is a scalar)

On comparing the coefficients of LHS and RHS,

-λ = 3

λ = 3

λ = -3

The value of λ is different

Therefore, points A, B, C are not collinear.

### Question 2 (ii) If  are non-coplanar vectors, prove that the points having the position vectors  are collinear.

Solution:

Let us assume three points that are A, B, C

Position vector of A =

Position vector of B =

Position vector of C =

= Position vector of B – Position vector of A

= Position vector of C – Position vector of B

By using  and , we get

= 2

So,  ||  but  is a common vector.

Hence, A, B, C are collinear

### Question 3. Prove that the points having position vectors  are collinear.

Solution:

Let us considered points A, B, C

Position vector of A =

Position vector of B =

Position vector of C =

= Position vector of B – Position vector of A

= Position vector of C – Position vector of B

By using and , we get

= -3

||  but  is a common vector.

Hence, A, B, C are collinear

### Question 4. If the points with position vectors  are collinear, find the value of a.

Solution:

Let the points be A, B, C

Position vector of A =

Position vector of B =

Position vector of C =

Given that, A, B, C are collinear

⇒  and   are collinear

⇒  = λ                     -(where λ is same scalar)

⇒ Position vector of B – Position vector of A = λ – (Position vector of C – Position vector of B)

On comparing the coefficients of LHS and RHS, we get

λa – 12λ = 2       -(1)

-8 = 11λ + 5λ       -(2)

-8 = 16λ

λ = -8/16

λ = -1/2

Now, Put the value of λ in eq(1), we get

λa – 12λ = 2

(-1/2)a – 12(-1/2) = 2

-a/2 +6 = 2

-a/2 = 2 – 6

-a/2 = -4

a = 8

So, the value of a is 8.

### Question 5. If  are two non-collinear vectors, prove that the points with position vectors  are collinear for all real values of λ.

Solution:

Let us considered points A, B, C

Position vector of A =

Position vector of B =

Position vector of C =

= Position vector of B – Position vector of A

= Position vector of C – Position vector of B

Using  and

=

Let  = μ

Since λ is a real number. So, μ is also a real number.

||  but  is a common vector.

Hence, A, B, C are collinear.

### Question 6. If , prove that A, B, C are collinear points

Solution:

According to the question

So,  ||  but  is a common vector.

Hence, A, B, C are collinear.

### Question 7. Show that the vectors  are collinear.

Solution:

Le us considered, the position vector A =

Position vector B =

Let us assume O be the initial point having position vector

= Position vector of A – Position vector of O

= Position vector of B – Position vector of O

By u8sing OA and OB, we get

Therefore,  ||  but O is the common point to them.

Hence, A and B are collinear.

### Question 8. If the points A(m, -1), B(2, 1), C(4, 5) are collinear, find the value of m.

Solution:

Let us considered

A = (m, -1)

B = (2, 1)

C = (4, 5)

= Position vector of B – Position vector of A

= Position vector of C – Position vector of B

A, B, C are collinear.

So,  and  are collinear.

So,  = λ

On comparing the coefficient of LHS and RHS, we get

2 – m = 2λ

-(1)

2 = 4λ

= λ

= λ         -(2)

From eq(1) and (2), we get

4 – 2m = 2

-2m = 2 – 4

m =

m = 1

So, the value of m is 1

### Question 9. Show that the points (3, 4), (-5, 16), (5, 1) are collinear.

Solution:

Let su considered

A = (3, 4)

B = (-5, 16)

C = (5, 1)

= Position vector of B – Position vector of A

= Position vector of C – Position vector of B

So,

||   but B is a common point.

Hence, A, B, C are collinear.

### Question 10. If the vectors  are collinear, find the value of m.

Solution:

Given:  and  are collinear.

So, a = λb

On comparing the coefficients of LHS and RHS, we get

2 = -6λ

λ = 2/(-6)

λ = -1/3          -(1)

-3 = λm

λ = -3/m          -(2)

From eq(1) and (2), we have

-1/3 = -3/m

m = 3 × 3

m = 9

So, the value of m is 9

### Question 11. Show that the points A(1, -2, -8), B(5, 0, -2) and C(11, 3, 7) are collinear, and find the ratio in which B divides AC.

Solution:

Given: A(1, -2, -8), B(5, 0, -2) and C(11, 3, 7).

So the given points are collinear

Now, let us considered point B divide AC in the ratio λ : 1.

Then

5(λ + 1) = 11λ + 1

⇒ 5λ + 5 = 11λ + 1

⇒ 6λ = 4

⇒ λ =

So, the point B divides AC in the ratio 2 : 3.

### Question 12. Using vector show that the points A(-2, 3, 5), B(7, 0, -1) and C(-3, -2, -5) and D(3, 4, 7) are such that AB and CD intersect at the point P(1, 2, 3).

Solution:

According to the question, we have,

= Position vector of P – Position vector of A

= Position vector of B – Position vector of P

Hence, . So, the vectors are collinear.

But P is a point common so, P, A, B are collinear points.

Similarly,

are collinear,

But P is a common point to . So, C, P, D are collinear points.

Hence, AB and CD intersect at the point P.

### Question 13. Using vectors, find the value of I such that the points (I, -10, 3), (1, -1, 3), and (3, 5, 3) are collinear.

Solution:

Given: Points (I, -10, 3), (1, -1, 3) and (3, 5, 3) are collinear.

Therefore, (I, -10, 3) = x(1, -1, 3) + y(3, 5, 3) for some scalars x and y

I = x + 3y          -(1)

-10 = -x + 5y          -(2)

3 = 3x + 3y          -(3)

On solving eq(2) and (3), we get,

x = 5/2 and y = -3/2

Now, put the value of x and y in eq(1), we get

I = (5/2) + 3(-3/2)

I = -2

So the value of I is -2

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